Question

In a sample of 41 temperature readings taken from the freezer of a restaurant, the mean is 29.7 degrees and the population standard deviation is 2.7 degrees. What would be the 80% confidence interval for the temperatures in the freezer?

Answer 1

Answer: $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

Step-by-step explanation:

Given: Sample size: n = 41

Sample mean $\overline{x}= 29.7$ degrees

Population standard deviation $\sigma=2.7$ degrees

Confidence level (c) =  80% =0.80

Significance level (a) = 1- c = 1-0.80 = 0.20

z-score for 80% confidence level : z = 1.2816   [from z-table]

Confidence level for population mean :-

$\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{\sqrt{41}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{6.403124}$

$=29.7\pm ( 1.2816)(0.42167)$

$=29.7\pm 0.5404$

$=(29.7-0.5404,\ 29.7+0.5404)$

$=(29.1596,\ 30.2404)$

Hence, 80% confidence interval for the temperatures in the freezer $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$